Show that there are infinitely many positive odd integers n such that n5+2n+1 is expressible as a sum of squares of two coprime integers.
Problem
Source: 2024 Korea winter program practice test P4
Tags: number theory
01.02.2024 20:06
Redacted
01.02.2024 20:14
I'm not sure I understand. Are you claiming that n5−n2 is infinitely often a perfect square? Or what excactly is the relation between the elliptic curve and the original problem?
01.02.2024 20:19
Redacted, nvm i was brain dead when I wrote this.
01.02.2024 20:59
I understand that y3=x2+1 has infinitely many rational solutions. But I still don't understand how this is related to the problem. Again, if you are claiming that n5−n2 is infinitely often a perfect square, this is just plain wrong.
03.02.2024 03:59
Some possible approach: n5+2n+1=a2+b2, (a,b)=1, n=2k+1 ⟹n5+2n+1=(2k+1)5+2(2k+1)+1=a2+b2 all odd prime divisors of a2+b2 have to be congruent to 1mod \Longrightarrow \frac{(2k+1)^5+2(2k+1)+1}{2}=\frac{32k^5+80k^4+80k^3+40k^2+14k+4}{2}= 16k^5+40k^4+40k^3+20k^2+7k+2=4M+1 \Longrightarrow 7k+2=1 \bmod {4}\Longrightarrow k=4t+1\Longrightarrow n=2k+1=2(4t+1)+1=8t+3 t=0 \longrightarrow n=3\longrightarrow \frac{3^5+6+1}{2}=125=2^2+11^2, ~~(2,11)=1. t=1 \longrightarrow n=11\longrightarrow \frac{11^5+22+1}{2}=80537=116^2+259^2,~~ (116,259)=1. t=3 \longrightarrow n=27\longrightarrow \frac{27^5+54+1}{2}=7174481=1779^2+2020^2, ~~(1779,2020)=1. t=2, 5 \longrightarrow n=19, 43\longrightarrow do not work (presence of two "4s+3" prime factors) Although it cannot be taken for granted that an even number of "4s+3" prime factors would not turn up, most cases work.
13.11.2024 09:12
Now that almost a year has passed, here's the official solution: Write 4(n+2)(n^5+2n+1)=(2n^3+2n^2-n+1)^2+(3n^2+22n+7). Since 3n^2+22n+7 is a square for infinitely many odd integers, we are done. (Using the theory of Pell equations we just need to check there is indeed a solution, and hopefully n=3 works. A little more work is needed to verify that (2n^3+2n^2-n+1, 3n^2+22n+7)=1 for these values.) \square The motivation behind this approach is that we want to make one term as a square of a polynomial, and the rest with low degree terms. The first thing that comes to mind to eliminate n^5 is to let n be a square itself, but sadly for that case 2n+1\equiv 3\pmod{4}. Twisting this and letting n be a monic quadratic fails because the best you can get for the remaining term is of degree 4. This gives insight that for this idea to succeed, one need to make a polynomial of degree at most 6. Now that there is only degree 1 that we can add from the starting polynomial, we figure that multiplying a linear term is our best attempt. (This is compatible to the fact that if AB is a sum of two coprime squares, then so are A and B.) On a side note, there was another beautiful solution by one of the students that exploits the fact that the linear term "looks" like it comes from n^2+2n+1. The sketch of the proof is to write n^5+2n+1=n^2(n^3-a^2n^2-2an-2a-1)+(an^2+n+1)^2, and find infinitely many (n,a) such n^3-a^2n^2-2an-2a-1 is a square.
13.11.2024 12:20
I remember that some problems from the AoPS user kaede has the same solution. Maybe the proposer was aware of it?
15.11.2024 10:52
Show that there are infinitely many positive odd integers n such that n^5+2n+1 is expressible as a sum of squares of two coprime integers: Doesn’t n =(2k+1)^2 -2 satisfy this? Then n^5+2n+1=[((2k+1)^2 -2)^2(2k+1)]^2 + 1^2 Or am I making some silly mistake?